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How do you find maclaurin series for #sinx cosx# using maclaurin for #sinx#?

1 Answer

Jul 1, 2016

#sinx cosx = sum_{k=0}^{oo}2^{2k}(-1)^k(x^{2k+1})/((2k+1)!)#

Explanation:

We know
#sin(a+b)=sin(a)cos(b)+sin(b)cos(a)# if #a = b#
#sin(2a)=2sin(a)cos(a)# then
#sinx cosx = sin(2x)/2#

Supposing that

#sin x = sum_{k=0}^{oo}(-1)^k(x^{2k+1})/((2k+1)!)#

then

#sinx cosx = sum_{k=0}^{oo}2^{2k}(-1)^k(x^{2k+1})/((2k+1)!)#

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